Question

With the assumption of no slipping, determine the mass m of the block which must be placed on the top of the 6.5-kg cart in order that the system period be 0.66 s. What is the minimum coefficient of static friction for which the block will not slip relative to the cart if the cart is displaced 67 mm from the equilibrium position and released?

Answer 1

Answer:

The minimum coefficient of static friction should be 0.62.

Explanation:

Given that,

Mass of block = m

Mass of cart = 6.5 kg

Time period = 0.66 s

Displacement = 67 mm

We need to calculate the mass of block

Using formula of time period

$T=2\pi\times(\dfrac{m}{k})$

Put the value into the formula

$0.66=2\pi\times(\dfrac{m+6}{600})$

$m=\dfrac{0.66\times600}{4\pi^2}-6$

$m=4.03\ kg$

We need to calculate the maximum acceleration of SHM

Using formula of acceleration

$a_{max}=\omega^2 A$

Maximum force on mass 'm' is $m\omega^2 A$

Which is being provided by the force of friction between the mass and the cart.

$\mu_{s}mg \geq m\omega^2 A$

$\mu_{s}\geq \dfrac{\omega^2 A}{g}$

$\mu_{s} \geq (\dfrac{2\pi}{T})^2\times\dfrac{A}{g}$

Put the value into the formula

$\mu_{s} \geq (\dfrac{2\pi}{0.66})^2\times\dfrac{0.067}{9.8}$

$\mu_{s} \geq 0.62$

Hence, The minimum coefficient of static friction should be 0.62.